$\overline{AB}$ = $\sqrt{85}$ $\overline{BC} = {?}$ $A$ $C$ $B$ $\sqrt{85}$ $?$ $ \sin( \angle BAC ) = \frac{7\sqrt{85} }{85}, \cos( \angle BAC ) = \frac{6\sqrt{85} }{85}, \tan( \angle BAC ) = \dfrac{7}{6}$
Solution: $\overline{AB}$ is the hypotenuse $\overline{BC}$ is opposite to $\angle BAC$ SOH CAH TOA We know the hypotenuse and need to solve for the opposite side so we can use the sine function (SOH) $ \sin( \angle BAC ) = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{\overline{BC}}{\overline{AB}}= \frac{\overline{BC}}{\sqrt{85}} $ $ \overline{BC}=\sqrt{85} \cdot \sin( \angle BAC ) = \sqrt{85} \cdot \frac{7\sqrt{85} }{85} = 7$